# Setting the limits of an integral

• docnet
In summary, the problem involves finding the area of a sphere enclosed by a cone using the equation da = r/(1-r^2) dr dθ. The limits of integration for the area integral are r from 0 to cosθ and θ from 0 to 2pi. By taking into account the expansion factor of r/sqrt(1-r^2) and dividing it by the expansion factor of 1/sqrt(1-r^2), we can find the correct area of pi^2 / 8 for the surface bounded by the intersection with the cone.
docnet
Gold Member
Homework Statement
Find the area of the unit sphere in R3 enclosed in the offset cone (x-1)^2 +y^2 = z^2
Relevant Equations
x^2 + y^2 + z^2 = 1

(x-1)^2 +y^2 = z^2
Problem: The sphere is parametrized in cylindrical coordinates by:

x = r cosθ
y = r sinθ
z = (1-r^2)^1/2

and intersected by the cone (x-1)^2 +y^2 = z^2.

find the area of the sphere enclosed by the cone using the equation:

da = r/(1-r^2) dr dθ
Attempt at solution:

from the equations for the sphere and cone: r = cosθ describes the intersection in the relevant coordinates.

the values of r ranges from 0 to 1, and θ from -pi/2 to pi/2.

How does one set the limits of integration for the area integral using da = r/(1-r^2) dr dθ ?

I tried setting r from 0 to cosθ and kept getting pi as the area, which is too large.

Delta2
We naturally imagine a map where the area enclosed by the graph of r = cosΘ in polar coordinates is ambiently popped up (projected) onto the surface of the sphere. The induced metric in polar coordinates gives da = r dr dθ. which means we take away the r in our da above and replace it with a 1. This is because the area is multiplied by a factor of 1/(1-r^2)^.5 under this map from polar to cylindrical coordinates, so the factor of expansion approaches infinity as r goes to 1, and 1 as r goes to 0. This makes sense because on the sphere , the slope of the the surface is infinite at the equator and 0 at the pole. so the amount of "stretching" varies accordingly.

integrating 1/(1-r^2)^.5 using the fore-mentioned limits gives pi^2 / 8 as the area of the surface bounded by the intersection with the cone.

Last edited:
Delta2
docnet said:
I tried setting r from 0 to cosθ and kept getting pi as the area, which is too large.
##r## going from ##0## to ##\cos\theta## is correct. I get ##\pi - 2## for the answer.

Delta2
LCKurtz said:
##r## going from ##0## to ##\cos\theta## is correct. I get ##\pi - 2## for the answer.

Oops, I am sorry. I made a typo in the original post and did not realize it until now. It should be r/sqrt(1-r^2). We obtained it by the following method.

In the previous part of the problem, we computed the expansion factor r/sqrt(1-r^2) by taking the square root of the determinant of the metric induced by the Euclian metric dx^2 + dy^2 + dz^2 in cylindrical coordinates by this map.

x = r cosθ
y = r sinθ
z = (1-r^2)^1/2

Computing it requires taking the partial derivatives of x, y, z and expanding it out. It simplifies to

(rdr)^2 + 1/(1-r^2) dθ^2

so the matrix of the metric is

r^2 0
0 1/(1-r^2)

Whose square root of the determinant is r/sqrt(1-r^2).Computing ∫∫ r/sqrt(1-r^2) dr dθ gives π which is too large. Solving this problem requires us to realize the expansion factor r/sqrt(1-r^2) needs to be changed. We have to divide this expansion factor by the other expansion factor corresponding to the metric induced by the Euclidian metric on polar coordinates.

r dr dθ

so we have

∫∫ 1/sqrt(1-r^2) dr dθ = pi^2 / 8
with
r: [0, cosθ]
θ : [
0, 2pi]

which is the surface area of the sphere enclosed within r = cosθ in cylindrical coordinates.

Last edited:
Delta2

## 1. What is the purpose of setting the limits of an integral?

The limits of an integral define the boundaries within which the function is being integrated. They determine the range over which the function will be evaluated and the area under the curve will be calculated.

## 2. How do you determine the limits of an integral?

The limits of an integral are determined by the specific problem or scenario being analyzed. They can be found by identifying the start and end points of the function being integrated, such as the x and y intercepts or points of intersection with other functions.

## 3. Can the limits of an integral be negative?

Yes, the limits of an integral can be negative. This is often the case when integrating over a region below the x-axis or when dealing with functions that have negative values.

## 4. What happens if the limits of an integral are not set correctly?

If the limits of an integral are not set correctly, the resulting calculation will be incorrect. This can lead to inaccurate solutions and incorrect interpretations of the problem being analyzed.

## 5. Is it possible to change the limits of an integral?

Yes, it is possible to change the limits of an integral. This can be done by using a change of variables or by using different methods of integration, such as substitution or integration by parts.

• Calculus and Beyond Homework Help
Replies
3
Views
619
• Calculus and Beyond Homework Help
Replies
9
Views
282
• Calculus and Beyond Homework Help
Replies
3
Views
411
• Calculus and Beyond Homework Help
Replies
8
Views
918
• Calculus and Beyond Homework Help
Replies
6
Views
997
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
986
• Calculus and Beyond Homework Help
Replies
14
Views
379
• Calculus and Beyond Homework Help
Replies
2
Views
843
• Calculus and Beyond Homework Help
Replies
21
Views
2K